Question

factorise (5a-4b)^3+(4b-3c)^3-(5a-3c)^3
be fast with steps​​

Answer 1

Step-by-step explanation:

Given :-

(5a-4b)³+(4b-3c)³-(5a-3c)³

To find:-

Factorise the expression?

Solution :-

Given expression is (5a-4b)³+(4b-3c)³-(5a-3c)³

It can be written as

=> (5a-4b)³+(4b-3c)³+[-(5a-3c)]³

=> (5a-4b)³+(4b-3c)³+(-5a+3c)³

It is in the form of x³+y³+z³

Where , x = 5a-4b , y = 4b-3c and

z = 3c-5a

And

x+y+z => 5a-4b+4b-3c+3c-5a = 0

We know that

If x+y+z = 0 then x³+y³+z³ = 3xyz

We have ,(5a-4b)³+(4b-3c)³+(-5a+3c)³

=> 3(5a-4b)(4b-3c)(-5a+3c)

=> 3(5a-4b)(4b-3c)(3c-5a)

Therefore, (5a-4b)³+(4b-3c)³-(5a-3c)³

= 3(5a-4b)(4b-3c)(3c-5a)

Answer:-

The factorization of (5a-4b)³+(4b-3c)³-(5a-3c)³ is

3(5a-4b)(4b-3c)(3c-5a)

Used formulae:-

• If x+y+z = 0 then x³+y³+z³ = 3xyz
• -(a)³ = (-a)³ = -a³

Answer 2

Answer:

$</p><p>[tex] \bold{ \mathtt{ {(5a - 4b)}^{3} + {(4b - 3c)}^{3} + {(3c - 5a)}^{3}}}$

$\bold{ \mathtt{ \blue{put \: \: \: 5a - 4b = p}}} \\ \bold{ \mathtt{ \blue{ \: \: \: \: \: \: \: \: \: 4b - 3c = q}}} \\ \bold{ \mathtt{ \blue{ \: \: \: \: \: \: \: \: \: 3c - 5a = r}}} \\ \\ \bold{ \mathtt{ \blue{now \: \: \: \: \: \: \: p + q + r}}} \\ \bold{ \mathtt{ \blue{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 5a - 4b + 4b - 3c + 3c - 5a}}} \\ \bold{ \mathtt{ \blue{ = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}}$

$\bold{ \mathtt{ \green{{(5a - 4b)}^{3} + {(4b - 3c)}^{3} + {(3c - 5a)}^{3}}}} \\ \bold{ \mathtt{ \green{ = {p}^{3} + {q}^{3} + {r}^{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}} \\ \bold{ \mathtt{ \green{ = 3pqr \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}} \\ \bold{ \mathtt{ \green{ = 3(5a - 4b)(4b - 3c)(3c - 5a)}}}$[/tex]